Question

A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it
The center of gravity of the beam is 2.00 m down the beam from the pivot. Lighting equipment exerts a 5.00-kN downward force on the lower left end of the beam. Find the tension T in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam.

Answer 1

Answer:

$T = 7.64 kN$

$F_y = 0.52 kN$(Downwards)

$F_x = 3.23 kN$ (Towards Left)

Explanation:

As we know that beam is in equilibrium

So here we can use torque balance as well as force balance for the beam

Now by torque balance equation at the pivot we can say

$F(4.50 cos\theta) + mg(2cos\theta) = T \times 3$

As we know that

mg = 1.40 kN

F = 5 kN

so we will have

$5 kN(4.50 cos25) + 1.40 kN(2 cos25) = 3 T$

$T = 7.64 kN$

Now force balance in vertical direction

$F + mg = Tsin65 + F_y$

$5 + 1.40 = 7.64 sin65 + F_y$

$F_y = 0.52 kN$(Downwards)

Force balance in horizontal direction

$F_x = T cos65$

$F_x = 7.64 cos65$

$F_x = 3.23 kN$ (Towards Left)